A hockey puck with mass 0.160 kg is at rest at the origin (x=0) on the horizontal, frictionless surface of the rink. At time t = 0 a player

Question

A hockey puck with mass 0.160 kg is at rest at the origin (x=0) on the horizontal, frictionless surface of the rink. At time t = 0 a player applies a force of 0.250 N to the puck, parallel to the x-axis; he continues to apply this force until t = 2.00s.

What is the position of the puck at t = 2.00s ?
In this case what is the speed of the puck?
If the same force is again applied at t = 5.00s , what is the position of the puck at t = 7.00s ?
In this case what is the speed of the puck?

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niczorrrr 6 months 2021-08-27T21:13:51+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-08-27T21:15:15+00:00

    Answer:

    a) x_{f} = 3.126\,m, b) v_{f} = 3.126\,\frac{m}{s}, c) x_{f} = 21.882\,m, d) v_{f} = 6.252\,\frac{m}{s}

    Explanation:

    a) Position of the puck:

    The acceleration experimented by the puck is:

    a_{puck} = \frac{0.25\,N}{0.160\,\frac{m}{s^{2}} }

    a_{puck} = 1.563\,\frac{m}{s^{2}}

    As the force remains constant during its time of application, accelaration is also constant. Position at given time is the following:

    x_{f} = \frac{1}{2}\cdot (1.563\,\frac{m}{s^{2}} )\cdot (2\,s)^{2}

    x_{f} = 3.126\,m

    b) Speed of the puck:

    The speed of the puck is computed as follows:

    v_{f} = (1.563\,\frac{m}{s^{2}} )\cdot (2\,s)

    v_{f} = 3.126\,\frac{m}{s}

    c) The absence of external force and the fact that ground is frictionless lead to the conclusion that hockey puck moves out at constant speed from 2 s. to 5 s. Then, the initial speed is:

    x_{o} = 3.126\,m + (3.126\,\frac{m}{s} )\cdot (5\,s-2\,s)

    x_{o} = 12.504\,m

    Likewise, the initial speed is 3.126\,\frac{m}{s}.

    The new application of the same force means the return of a accelerated movement. Then:

    x_{f} = 12.504\,m+(3.126\,\frac{m}{s} )\cdot (7\,s-5\,s)+\frac{1}{2}\cdot (1.563\,\frac{m}{s^{2}} )\cdot (7\,s-5\,s)^{2}

    x_{f} = 21.882\,m

    d) The final speed of the puck is:

    v_{f} = 3.126\,\frac{m}{s} + (1.563\,\frac{m}{s^{2}})\cdot (7\,s-5\,s)

    v_{f} = 6.252\,\frac{m}{s}

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